Ok so what we need to do is:
p = 312/624 = 0.500 best estimate
mean is p N = (0.500) 624 = 312
standard deviation is sqrt[Np(1-p)] = 12.5
If we approximate the binomial by a Normal
we would be using
mu = 312, s.d = 12.5
and the 99% C.I. is
mean +- 2.57 s.d
we convert the mean to p by dividing by 624
so corresponding C.I. for p would be 0.500+ - (2.57)(12.5)/624
or 0.500+- 0.051
I hope this is useful for you